189
Internat. J. Math. & Math. Sci.
(1998) 189-196
VOL. 21 NO.
RESEARCH NOTES
THE SOLUTION OF A SINGULAR INTEGRAL EQUATION WITH SOME APPLICATIONS
IN POTENTIAL THEORY
N.T. SHAWAGFEH
Department of mathematics
University of Jordan
Amman, JORDAN
(Received January 6, 1995 and in revised form December 9, 1996)
ABSTRACT. An analytical solution is derived for a singular integral equation which governs some twodimensional potential boundary value problems in a region exterior to n-infinite co-axial circular strips An
application in electrostatics is discussed
KEY WORDS AND PHRASES:Singular integral equation,electrostatics, Laplace’s equation
1991 AMS SUBJECT CLASSIFICATION CODES: 31Bl0, 45B05, 78A30
INTRODUCTION
In this paper we derive a solution to the Fredhoim singular integral equation
where q and a are constants and I"(0)is a differentiable nction for
]0- (2k# n)l < a ,k
0
n-I This
integral equation governs the solution of various two-dimensional Difichlit and Neumann potential
bounda value problems for the region consisting of the whole (r, O) plane outside the circular strips
2k
The previous investigations in potential problems of circular strips [1-8] were concerned mainly with
the case of two strips, where Greens nction approach leads to a singular integral equation with kernel
q + log
sin(O-)
Shail [3] transfoed this equation into a well known Cademan type
A different technique has been used by Sampath and Jain [4] based on decoupling the equation into
two singular equations which can be solved using eigennctions expansion metod The same technique
has been used to solve various bounda value problems involving two circular strips[ 5-8
N. T. SHAWAGFEH
190
In section 2 we use the approach of[4] to solve equation(1, l),and in section 3 we apply the results to
solve laplace equation associated with Dirichlit and Neumann boundary conditions As an illustrative
example we consider in section 4 the electrostatics problem in a region external to perfectly conducting
n- circular strips Formulae for the surface charge density and concentration factor are derived,and these
are believed to be new
2.SOLUTION OF THE INTEGRAL EQUATION
To obtain the solution of the integral equation (1 l) we first reduce it to simpler singular integral
equations, so interchanging the sum and integral signs and using some properties of the function in the
kernel we obtain
1()
Using the identity [9],
(] cos(Oin the equation (2
we get
g’j(lcs(O---t_0
-
qn--iog2 + 2
2r))
()[Q + og(1
where (.)
J
--;-i-(l- cos(n(O- ))),
(2.2)
I<a
cos,,(8- )]d 2F(a),
(2 3)
2qn (2n l) log 2
Now we write each of the functions l(0)and F(0) as
l(O)
ie(8)+ lo(e
(2 4a)
l,’(O)
I., (/9) + I, (O)
(2 4b)
where the subscripts e and o stand for the even and odd parts respectively. Substituting (2 4) into (2 3)
we obtain the following decoupled singular integral equations
/,,(#)log
n
0<8 <a
(2 5)
O<O <or
(2 6)
2
in order to solve eq (2 5) we introduce the transformation
cos(,,)
cos(n#)
sin:()cosx + cos:(),
sin
()cosy + cos ()
O< x <n"
(2 7a)
O< y <
(2 7b)
n"
which transforms the equation into the form
j H(y)[ R + log(21cosx
where
cosyiy
h(x),
O< x <r
(2 8)
191
SOLUTION OF A SINGULAR INTEGRAL EQUATION
R=Q+log
sin z(,,a/2)
(29)
(2 10)
sin" (-2-) sin y
and H(y)=
nsin(n#)
Expanding h(x) as Fourier cosine series
where
.
-a +
h(x)
(2 I1)
I())
cos(rex)
O<X<K
(2 12)
m>_O
(2 13)
[ h(x) cos(mx)dx.
a.
and making use of the bilinear form 10]
log( 21cos x cos Yl)
cos( mx) cos(my)
-2
(2 14)
O<x, y
m
we find that the solution of (2 8) can be written in the form
H(y)
4,, +
a,,
A..
(2
,4. cos(my),
where
A,,
2M?
m>
---a,.
r
(2 16)
thus using (2 11 we find that the solution of (2 5) is given by
!,,()
.-ncos(n 2)
/cos(,,)- cos(,,a)
A +
A., cos(my)
(2 17)
O<
where we have used the relation
sioo.
Csc:
(2 18)
Now to solve eq (2 6) we first differentiate both sides with respect to8
nlo() sin(nO0
d b"(e)
cos(,,)-cos(-O)
(2 19)
Applying the transformation (2 7) into (2 19) reduces it to
G(y)
dy
cosy cos x
O<x <
g(x).
(2 20)
where
g(x)
G(y)
1."(o)
(221)
1o(#)sin y
(2 22)
Expanding (;(y)in Fourier cosine series
(;(y)
B,, +
B.. cos(my).
O<y< zr
(2 23)
N. T. SHAWAGFEH
192
and noting the integral
cos(my)
dr’= zcsin(mx)csc(x)
cos)’ cos x
which follows by putting w= e" and integrating around the circle
G(y)
dy
cosy COS X
O<X <
]w[
7
(224)
Then
(225)
B, sin(mx)csc(x),
Thus if we expand
(’, si n(mx)
(226)
g(x)sinxsin(mx)dx
(2.27)
g(x) sin x
it follows that
m_>l
(228)
(" cos(my)
(2 29)
and hence
I,,() _ml
slrly
B,, +
zr
To evaluate the constant B,, we use the finiteness condition that G(y) --, 0 as y
-
0 this gives
(230)
Substituting (2 30) into (2 29) we find from (2 22) and the relation (2 18) that
/(0)
(’(1-cosmy)
siny.,
-a
< <a
(231)
Finally summing up the results we write the solution of eq (1 l) as
/()
cosO,)- cos(,,a)
nsiny
(’= (1- cosmy)
-a <
<a
(232)
where
(233)
4n’[qn + log(2
m>l
m>O
h( x cos(mx)dx,
a.,
("’
=- re2 !
h(x)
!.i,(0
g(x) sin x sin(mx)dx
g(x)
!,i;(0
(2 34)
(235)
(2 36)
(2 37)
193
SOLUTION OF A SINGULAR INTEGRAL EQUATION
3.APPLICATION IN POTENTIAL THEORY
The singular integral eq (11) governs the solutions of various Dirichlit and Neumann problems for
two-dimensional Laplace equation in the domain D consisting of all the points (r,0)lying ooutside the ncircular infinite strips
( r=a,
[0-zk
(3 I)
<a,k 0 n-l,
These types of problems appear in different physical fields such as electrostatics, magnetostatics, steady
state heat flow, and others
A Dirichlit problem We seek a function (1)(r, 0)that satisfies the foilwing boundary value problem
V:(1)
(32)
0
lo_2krl<a
(1)(a,O): f(O)
(l)(r,
!-"
0)1------ logr
k 0
n_ 1
(33)
(3 4)
r --->
and are continuous across the arcs
(’;
2k;r
+a
r=a
r
<0< 2(k+l)---at
(35)
ll
--
The function ./’(0) satisfies the symmetry condition
.[(o 2kn’l
+
11
,
(36)
-a < O < at
./(O)
./
c,
(37)
ds
Using Green’s function approach and the symmetry in the problem the solution can be represented as
a i/r.., +a"-2racos(O-b-2kZC)dqb,,
*(r. O)
i (,)log
(3 8)
where the density funcUon l(b) is defined as
The boundary condition (3 3) leads to the following integral equation
(3
which is eq (1 1) with q
log(2a)
F(O)
-2" f(O), and hence it has the solution (2 32) with the
a
substitution of these values
B. Neumann problem We seek a function (r, O) satisfying the boundary value problem
V2T =0
T(a,O)= p(O)
---<at
,k 0 n-I
(312)
N. T. SHAWAGFEH
194
q(r,O)-O ,as r
are continuous across the arcs
and
-
(3 13)
The function I’(0) satisfies the symmetry condition
p(o 2kzCl
p(O)
+
-ct
(3 14)
<O <
and the consistency condition
i p(O)dO
(3 15)
0
Green’s function approach leads to the follwing representation
W(r,O)
--N ,.](#)
log
r: +/9:
2kzt)
2r,ocos(0- #-
II
d#
-a.
where./()=[V(r,]:
(3 16)
(3 17)
Fuhermore .1() satisfies the edge condition
-
(3 8)
./(a) 0
Using the bounda condition (3 12) we obtain
’/(’)csc: 7(02
2k)d’
(3 19)
80)
Integrating by pans using the edge condition (3 18) we obtain
J
(#)cot(0-#
Upon integration with respect to
F(O)
(3 20)
-2’(0) +
(321)
we get
,/’()oglsin(O- #where 1’()
4N0)
/1.
,, ) d#
p(O)dO and (" is an arbitra constant Eq (3 21)is speci se ofeq (1 1) with q
(’- 2p(O), and
I(#) .1’(#) To deteine the constant
0
(’we use the edge condition (3 18)
Finally the density .1() is given by
.I’()d#
.1(0)
a
Oa
(3 22)
4.1LLUSTTIVE EXAMPLE
As an example to illustrate our results we consider the electostatics potenti problem of the n-equal
infinite co-axial perfectly conducting strips in flee space charged so that thetotal charge per height on
each strip is unity.ln cylindrical coordinates
r=a,
O-
-
(r, 0,z)It the stips be defined by
<a
,
0
n-l,,
(41)
SOLUTION OF A SINGULAR INTEGRAL EQUATION
195
then the electrostatics potential satisfies the Dirichlit problem A, where on the boundary we assume that
has the constant potential K Thus
(1)(r. 0)
and l(q)
-a"’i (q) log [r
+ a’-
2ra cos(0
.
q
2krC)dq
(4 2)
satist the integral equation (3 10) with f(0)=K ,and hence
#(x)
I, (0)=
-K
,
g(x)
(o) o
(4 3)
Substituting these valus in (2 32) we find that
K, cos(n 2
(4 4)
To find the value of K we use the condition that the total charge per unit height on each strip is unity.that
is.
(4 5)
Inseing (4 4) into (4 5) and evaluating the resulting integral wefind that
K -tog"}sin ,,a 2I)
(4 7)
and the charge density in this case is
/(0)
, cos(n
2)
’-a
m$co,0)- cos(-a)
<0 < a
(48)
while the charge concentration factor is given by
N
limCa -0)
’
<0)
na
4ncot(
2a
2)
(4 9)
All the above formulae are believed to be new For n=4 the formulae (4 6).(4 7).and (4 8) agree with
those in [5].
and when a
we recover the known results for a charged infinite cylinder when the total charge
per unit height is n
ACKNOWLEDGMENT. The present work is supported by the university of Jordan under project
#356
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N. T. SHAWAGFEH.
196
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